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91Ó°ÊÓ

Calculate the binding energy per nucleon for the following nuclei: (a) \({ }_{6}^{12} \mathrm{C}\) (nuclear mass, \(11.996708\) amu); (b) \({ }^{37} \mathrm{Cl}\) (nuclear mass, \(36.956576\) amu ; (c) rhodium-103 (atomic mass, \(102.905504\) amu).

Short Answer

Expert verified
For each nucleus, the binding energy per nucleon is calculated as follows: a) Carbon-12: - Mass defect: (11.996708 amu) - [(6 * 1.007276466812 amu) + (6 * 1.008664915 amu)] = 0.0989312 amu - Convert to energy: E = (0.0989312 amu * 1.66053906660 * 10^{-27} kg/amu) * (2.998*10^8 m/s)^2 ≈ 1.633860 * 10^{-13} J - Binding energy per nucleon: E/12 ≈ 1.36155 * 10^{-14} J/nucleon b) Chlorine-37: - Mass defect: (36.956576 amu) - [(17 * 1.007276466812 amu) + (20 * 1.008664915 amu)] = 0.235228989016 amu - Convert to energy: E = (0.235228989016 amu * 1.66053906660 * 10^{-27} kg/amu) * (2.998*10^8 m/s)^2 ≈ 3.922859 * 10^{-13} J - Binding energy per nucleon: E/37 ≈ 1.060971 * 10^{-14} J/nucleon c) Rhodium-103: - Mass defect: (102.905504 amu) - [(45 * 1.007276466812 amu) + (58 * 1.008664915 amu)] = 1.49022702704 amu - Convert to energy: E = (1.49022702704 amu * 1.66053906660 * 10^{-27} kg/amu) * (2.998*10^8 m/s)^2 ≈ 2.482038 * 10^{-12} J - Binding energy per nucleon: E/103 ≈ 2.409063 * 10^{-14} J/nucleon

Step by step solution

01

Calculate the mass defect

For each nucleus, we need to find the difference between the mass of the nucleus and the mass of its individual protons and neutrons. a) Carbon-12: - Mass of carbon-12 nucleus: 11.996708 amu - Mass of a proton: 1.007276466812 amu - Mass of a neutron: 1.008664915 amu - 6 protons and 6 neutrons: (6 * 1.007276466812 amu) + (6 *1.008664915 amu) b) Chlorine-37: - Mass of chlorine-37 nucleus: 36.956576 amu - 17 protons and 20 neutrons: (17 * 1.007276466812 amu) + (20 *1.008664915 amu) c) Rhodium-103: - Mass of rhodium-103 nucleus: 102.905504 amu - 45 protons and 58 neutrons: (45 * 1.007276466812 amu) + (58 *1.008664915 amu)
02

Convert the mass defect into energy

Now, we will use the Einstein relation to convert the mass defect into energy. E = mc^2 Where: - E is the energy - m is the mass defect - c is the speed of light (2.998*10^8 m/s) We will also convert the mass defect from atomic mass units (amu) to kilograms (kg) using the conversion factor: 1 amu = 1.66053906660 * 10^-27 kg.
03

Calculate binding energy per nucleon

Now that we have converted the mass defect to energy, we will divide it by the total number of nucleons (protons + neutrons) for each nucleus. a) Carbon-12: - Total nucleons: 12 - Binding energy per nucleon: E/12 b) Chlorine-37: - Total nucleons: 37 - Binding energy per nucleon: E/37 c) Rhodium-103: - Total nucleons: 103 - Binding energy per nucleon: E/103 Now the calculations for each nucleus will provide the respective binding energy per nucleon.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nuclear Mass
Understanding nuclear mass is fundamental to grasping the concept of binding energy per nucleon. Nuclear mass refers to the total mass of the protons and neutrons, collectively known as nucleons, within an atomic nucleus. When calculating the nuclear mass of an atom, you might expect it to be simply the sum of the masses of its protons and neutrons. However, due to the binding energy that holds the nucleus together, the actual nuclear mass is slightly less than this sum.

This discrepancy arises because energy has been released when the protons and neutrons bounded together to form the nucleus. The system loses mass in the form of energy, which is an example of the mass-energy equivalence principle proposed by Albert Einstein. In the context of the exercise, we see that precise values of nuclear mass are provided, such as 11.996708 atomic mass units (amu) for a Carbon-12 nucleus, which are critical for calculating the mass defect and subsequently the binding energy per nucleon.
Mass Defect
The concept of mass defect plays a pivotal role in understanding nuclear stability and energy. The mass defect is the difference between the calculated mass of the constituent protons and neutrons and the actual nuclear mass measured experimentally. It represents the amount of mass converted into energy during the formation of the nucleus. This energy is the binding energy, which as the name suggests, is the energy required to bind the nucleons together in the nucleus.

In the task's step by step solution, the mass defect is calculated by subtracting the nuclear mass from the combined mass of the individual protons and neutrons. For instance, for Carbon-12, multiplying the mass of a single proton and neutron by 6, adding them together, and then subtracting the nuclear mass gives the mass defect for the Carbon-12 nucleus. This mass defect, once calculated, can then be used to determine the total binding energy of the nucleus.
Einstein's Mass-Energy Equivalence
One of the most famous formulas in physics, Einstein's mass-energy equivalence equation, E = mc^2, expresses the idea that mass and energy are interchangeable. This principle is crucial in the realm of nuclear physics. In essence, the mass defect associated with a nucleus's binding energy is a manifestation of this equivalence.

In the exercise solution, the mass defect—in atomic mass units—is first converted to kilograms to comply with the units used in the mass-energy equivalence formula. The binding energy is then calculated using the speed of light squared (c^2), a constant value of approximately 2.998 x 10^8 meters per second. Understanding this equivalence allows us to appreciate how much energy is tied up within even the smallest amount of mass within an atom's nucleus, and it demonstrates the incredibly efficient 'storage' of energy at the nuclear level. As the lingering question of the exercise points out, the binding energy is allocated per nucleon to understand the stability of different nuclei, highlighting the power and utility of Einstein's groundbreaking relation in nuclear physics.

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Most popular questions from this chapter

An experiment was designed to determine whether an aquatic plant absorbed iodide ion from water. Iodine\(131\left(t_{1 / 2}=8.02\right.\) days) was added as a tracer, in the form of iodide ion, to a tank containing the plants. The initial activity of a \(1.00-\mu \mathrm{L}\) sample of the water was 214 counts per minute. After 30 days the level of activity in a \(1.00-\mu \mathrm{L}\) sample was \(15.7\) counts per minute. Did the plants absorb iodide from the water? Explain.

Americium-241 is an alpha emitter used in smoke detectors. The alpha radiation ionizes molecules in an air-filled gap between two electrodes in the smoke detector, leading to current. When smoke is present, the ionized molecules bind to smoke particles and the current decreases; when the current is reduced sufficiently, an alarm sounds. (a) Write the nuclear equation corresponding to the alpha decay of americium-241. (b) Why is an alpha emitter a better choice than a gamma emitter for a smoke detector? (c) In a commercial smoke detector, only \(0.2\) micrograms of americium are present. Calculate the energy that is equivalent to the mass loss of this amount of americium due to alpha radiation. The atomic mass of americium- 241 is \(241.056829\) amu. (d) The half-life of americium-241 is 432 years; the half life of americium- 240 is \(2.12\) days. Why is the 241 isotope a better choice for a smoke detector?

Predict the type of radioactive decay process for the following radionuclides: (a) \({ }_{5}^{8} \mathrm{~B}\), (b) \({ }_{29}^{68} \mathrm{Cu}\), (c) phosphorus32, (d) chlorine-39.

The naturally occurring radioactive decay series that begins with \({ }_{92}^{235} \mathrm{U}\) stops with formation of the stable \({ }_{82}^{20} \mathrm{~Pb}\) nucleus. The decays proceed through a series of alpha-particle and beta-particle emissions. How many of each type of emission are involved in this series?

The Sun radiates energy into space at the rate of \(3.9 \times 10^{26} \mathrm{~J} / \mathrm{s} .(\mathrm{a})\) Calculate the rate of mass loss from the Sun in \(\mathrm{kg} / \mathrm{s}\). (b) How does this mass loss arise? (c) It is estimated that the Sun contains \(9 \times 10^{56}\) free protons. How many protons per second are consumed in nuclear reactions in the Sun?

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